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Conditioning on $T_1=t$ and using independent random variables giveswhere the notation in the paper uses $F_2$ for the survival function.

Write the Cepheid measurement as $\hat{\mu }=\mu +{ϵ}_{\mu }$, where ${ϵ}_{\mu }\sim N(0,{\sigma }_{\mu }^2)$, and define the calibrated absolute magnitude dataThus $q=(q_1,\dots ,q_K{)}^T$ has a multivariate normal distribution with mean $M_{0}1$ and covariance matrix

The Hubble law and the definition of distance modulus giveConsequently, withthe Hubble-flow observations are independent random variables satisfying $x_i\sim N(M_0-\theta ,{\sigma }_{\mathrm{tot}}^2)$. Apart from a factor independent of the parameters, the joint likelihood function isThe expression extends by continuity to a singular limiting covariance such as ${\sigma }_I=0$.

Let $(X_n)$ be independent random variables and letbe their tail sigma-algebra. The Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Indeed, $A$ is independent of $\sigma (X_1,\dots ,X_n)$ for every $n$. The union of these finite-coordinate sigma-algebras generates $\sigma (X_1,X_2,\dots )$, and independence extended from generating pi-systems shows that $A$ is independent of that entire sigma-algebra. Since $A$ itself belongs to it,so $\mathbb{P}(A)\in \{0,1\}$.

A function has the bounded-differences property with constants
$L_1,\dots ,L_n$ when replacing only coordinate $i$ changes its value by at most $L_i$. The Bounded differences inequality states that, for independent inputs,and the analogous lower-tail bound also holds.

For an open subdomain $U\subset D$, identify $H_0^1(U)$ with the closed subspace of $H_0^1(D)$ obtained by zero extension. Its orthogonal complement isIf $f\in {\mathcal{H}}_U$, then testing against $C_0^{\infty }(U)$ and integrating by parts gives $\Delta f=0$ in $U$ in the distributional derivative sense. The Weyl lemma therefore gives a representative harmonic on $U$.

The orthogonal decomposition by a closed subspace givesProject the isonormal process defining $h$ onto these two orthogonal subspaces. The projections are jointly Gaussian and uncorrelated, hence independent. The first projection is a zero-boundary Gaussian free field $h_U$ on $U$; the second is a random distribution $h^{\mathrm{har}}$ that is harmonic on $U$. Thuswith independent summands. This proves the Domain Markov property of the Gaussian free field.

Because the variables are independent, the moment-generating function of the sum factors. Whenever $|\lambda |<1/{\max }_i{\alpha }_i$,Therefore ${\sum }_i{X}_i$ is sub-exponential with parameters

Let $X_1,\dots ,X_n$ be independent random variables, let $f=f(X_1,\dots ,X_n)$, and let each $f_i$ depend on every coordinate except $X_i$. The modified logarithmic Sobolev inequality states that, for every real $\lambda$ for which the expectations exist,

To prove it, first apply tensorization of entropy:Condition on all coordinates except $X_i$ and use the stated variational formula with the admissible constant $u=e^{\lambda f_i}$. The $i$th conditional entropy is at mostSumming and taking the remaining expectations proves the inequality.

Model 1's residuals have a systematic curved pattern and a spread that grows strongly with fitted time. This indicates an incorrect linear mean on the original scale and heteroscedasticity; several observations are also influential or outlying. The constant-variance assumption is therefore implausible.

Logging time greatly stabilizes the spread and removes most of the mean pattern, so model 2 is much more compatible with constant conditional variance and linearity. A few conspicuous residuals remain. A residual-versus-fitted plot alone does not check independence or fully establish normality.

Because $X$ and $Y$ are independent random variables, conditioning on $Y$ merely translates $X$. The conditional entropy under a deterministic change of variables and the fact that conditioning reduces entropy therefore giveInterchanging $X$ and $Y$ similarly gives $H(X+Y)\ge H(Y)$, and henceThe proof applies to countable alphabets whenever the displayed information entropies are finite.

Because a strictly monotone map fixing both endpoints is increasing, the change of variables formula $t=h(s)$ givesAssuming $\mathbb{E}\Vert X{\Vert }^2>0$, defineThe Hilbert-space variance identity then gives

Under the regularity needed to interchange expectation and differentiation, $\mathbb{E}{h}^{\prime }(s)=1$ because $\mathbb{E}h(s)=s$. HenceThus $A=1$ exactly when the integrated covariance in the numerator vanishes. In particular, this holds when the amplitude $X$ and the time warp $h$ are independent random variables.

A one-time pad over the finite additive group ${\mathbb{Z}}_n$ uses a key $K$ that is uniform on ${\mathbb{Z}}_n$, independent of the plaintext $X$, as long as the plaintext, and never reused. Encryption and decryption are

For independent random variables $X,Y$, conditioning reduces entropy and translation by a known element of the finite additive group preserves entropy, soInterchanging $X$ and $Y$ gives $H(X+Y)\ge H(Y)$, henceThis is the entropy of a sum of independent finite-group variables.

The result explains why adding independent pad symbols cannot reduce uncertainty. For a uniform pad, $X+K$ is itself uniform and is independent of $X$, which is the perfect-secrecy property. Independence is necessary: if $X$ is nonconstant and $Y=-X(\mathrm{mod}n)$, then $X+Y=0$ is constant, so

The commands choose $n=100$ observations and $p=2$ predictors; generate a $100\times 2$ matrix $X$ of independent standard-normal values; generate independent binary random variables $Y_i$ with success probability $1/2$; and report the observed number of successes, ${\sum }_i{Y}_i=48$. The first glm call then fits the Bernoulli logistic-regression modelThe prediction command returns the fitted response probabilities ${\hat{p}}_i$ and sums them.

The likelihood isAt the maximum-likelihood estimator, the intercept component of the score function isThereforeexactly, by the fitted-mean balance for logistic regression with an intercept.

The second fit uses probit regression,where $\Phi$ is the standard normal distribution function. Its intercept score isa weighted residual equation. It does not imply ${\sum }_i{\hat{p}}_i=48$, so exact equality is not expected. The output should nevertheless be close to $48$: the data were generated with constant success probability $1/2$ independently of $X$, so the fitted slopes should be small, the linear predictors should cluster near a common intercept, and the score weights should be nearly constant.

Let $X_t$ be the number of telephone lines in use. The memorylessness of the exponential distribution makes $(X_t{)}_{t\ge 0}$ a continuous-time Markov chain, specifically the M-M-infinity queue with generator rates

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the exponential survival probability $p_t$, independently of the others. Their surviving number is thereforeA call arriving at time $u\in [0,t]$ survives until $t$ with probability $e^{-\mu (t-u)}$. By Poisson thinning, the number of surviving new calls is independent of $Y_t$ and has lawHence $X_t=Y_t+Z_t$ in distribution. Multiplying the probability generating functions of the two independent random variables gives, for $-1\le s\le 1$,Equivalently, the transient distribution of an M-M-infinity queue iswith independent summands. As $t\to \infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda /\mu$. Thereforewhich proves the stated large-time approximation.

Let $Y_i$ be the first team's goal count in match $i$, let $x_i^T$ be row $i$ of the design matrix, and let $\beta$ be the regression coefficient vector. The fitted Poisson regression assumes that the $Y_i$ are independent random variables withThe likelihood function maximized by glm is therefore

Write $S_n={\sum }_{j=1}^n{Y}_j$ for the partial sum. Fix an integer $M\ge 1$. The tail-sum formula for expectation, applied to the nonnegative random variable $|Y_1|$, givesbecause $\mathbb{E}|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are independent, so the second Borel--Cantelli lemma shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in \mathbb{N}$ yieldsalmost surely.

Now $Y_n=S_n-S_{n-1}$, and the triangle inequality givesIf the limit superior of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Thereforealmost surely.

A Non-homogeneous Poisson point process with intensity function $\lambda$ is a random counting measure $N$ such that, for every finite family of pairwise disjoint bounded Borel sets $A_1,\dots ,A_k$, the counts $N(A_1),\dots ,N(A_k)$ are independent random variables andThe assumed local integrability makes each Poisson parameter finite. This count law and independence characterize the process.

They are not independent. The Itô formula gives ${\beta }_t^2=t+2{\int }_0^t{\beta }_{s}d{\beta }_s$, and the bilinear Itô isometry therefore givesIf $B_t$ and ${\beta }_t$ were independent random variables, then $B_t$ would also be independent of the measurable function ${\beta }_t^2$, and centeredness would instead give $\mathbb{E}[B_t{\beta }_t^2]=\mathbb{E}[B_t]\mathbb{E}[{\beta }_t^2]=0$. This contradiction disproves independence.

Write the molecular copy numbers as $\mathbf{x}=(x_1,x_2,x_3,x_4)$ and use the power-law reaction propensity convention of the paper. The seven propensities and stoichiometric vectors areA channel is assigned zero propensity whenever its update would leave the nonnegative integer lattice.

The Gillespie algorithm starts from $\mathbf{x}=(5,0,0,0)$ and $t=0$. At the current state compute $a_0={\sum }_{r=1}^7{a}_r$. If $a_0=0$, terminate the path. Otherwise draw independent random variables $U_1,U_2$ from the uniform distribution on $(0,1)$, set the next waiting time toand choose the least index $r$ satisfyingThen update $t\leftarrow t+\tau$ and $\mathbf{x}\leftarrow \mathbf{x}+{\nu }_r$, and repeat. The minimum of the seven competing reaction clocks has an exponential distribution of rate $a_0$, while reaction $r$ wins with probability $a_r/a_0$.

Let ${\xi }_{1,n},{\xi }_{2,n}$ be independent random variables with the standard normal distribution. The unconstrained Milstein method step isThe $X$ noise is additive, so its Milstein correction vanishes. The diagonal, independent noise fields commute, so no cross iterated stochastic integral is required.

Implement each reflecting boundary condition for a diffusion by folding the proposed coordinate back into $[-1,1]$. One formula that also handles multiple overshoots isThe complete update isUnder the usual smoothness assumptions, the Milstein discretization has strong order of convergence one and weak order of convergence one. The reflection enforces the boundary pathwise.

Because $X,Y,Z$ are independent random variables, adding $Y$ is an independent noise channel, sois a Markov chain. The data processing inequality for mutual information givesTranslation in the finite additive group preserves conditional entropy, and independence therefore givesandSubstitution proves the required entropy submodularity for three independent sums.